Given that the absolute value of 2A + B is opposite to 3A + B-1 under the root sign, find the value of bracket a + B to the power of 2010

Given that the absolute value of 2A + B is opposite to 3A + B-1 under the root sign, find the value of bracket a + B to the power of 2010

|2a+b|=0
3a+b-1=0
a=1 b=-2
(a+b)^2010=1

Given Tan α = 2, find 2 / 5cos ^ 2 α - 1 / 2Sin ^ 2 alpha

tanα=2
2/5cos²α-1/2sin²α/1
=(2 / 5cos? α - 1 / 2Sin? α) / (sin? α + cos? α) both sides divide cos? α
=(2/5-1/2tan²α)/(tan²α+1)
=-8/25

Solving inequality group x + 3 < 0.2 (x-1) + 3 ≥ 3x

x+3＜0
x=3x
-x>=-1
X

If Sina = 2 / 5 √ 5, and cosa

From Sina = 2 / 5 √ 5, and cosa

Evaluation: tan70 ° cos10 ° × (√ 3tan20 ° - 1)

tan70×cos10×(√3tan20-1) =tan70×cos10×(tan60×tan20-1) =tan70×cos10×[(sin60×sin20/cos60×cos20)-1] =tan70×cos10×(sin60×sin20-cos60×cos20)/(cos60×cos20) =tan70×cos10×[-cos(60+20)]/(cos60 ...

The value of sin10 ° sin30 ° sin50 ° sin70 ° is () A. 1 Two B. 1 Four C. 1 Eight D. 1 Sixteen

Original formula = 1
2•2sin10°cos10°cos20°cos40°
2cos10°＝sin80°
16cos10°＝1
Sixteen
Therefore, D

Calculation: cos 20 degrees * cos 40 degrees * cos 60 degrees * cos 80 degrees

Original formula = 2sin20? * cos20? * cos40? * cos60? * cos80? / (2sin20?)
=sin40º*cos40º*cos60º*cos80º/(2sin20º)
=sin80º*cos60º*cos80º/(4sin20º)
=sin160º*cos60º/(8sin20º)
=sin(180º-20º)/(16sin20º)
=sin20º/(16sin20º)
=1/16

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Multiply X-1 on both sides of equation 2, and x ^ 2-x + 1 = 7x-7
X ^ 2-8x + 8 = 0
X = 4-2 root sign 2,
The following do not understand your title, find x, you put in it

Without a computer, compare the size of root 7 plus root 5 and root 2 times root 6, and explain the reasons

(7+5^(1/2))^2=49+5+14*5^(1/2)=54+14*5^(1/2)
(2*6^(1/2))^2=24
So 7 + 5 ^ (1 / 2) > 2 * 6 ^ (1 / 2)

Find sin20 ° ^ 2 + cos80 ° ^ 2 + root 3sin20 ° cos80 ° Come on, it's a process

=(1-cos40)/2+(1+cos160)/2+sqrt(3)/2(sin100-sin60)
=1-1/2cos40-1/2cos20-3/4+sqrt(3)/2sin100
=1/4-1/2(cos40+cos20)+sqrt(3)/2sin80
=1/4-cos10cos30+sqrt(3)/2sin80
=1/4-sqrt(3)/2cos10+sqrt(3)/2sin80
=1/4